Question: Divide the following complex numbers. $ \dfrac{-26+7i}{-4+3i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-4-3i}$ $ \dfrac{-26+7i}{-4+3i} = \dfrac{-26+7i}{-4+3i} \cdot \dfrac{{-4-3i}}{{-4-3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-26+7i) \cdot (-4-3i)} {(-4+3i) \cdot (-4-3i)} = \dfrac{(-26+7i) \cdot (-4-3i)} {(-4)^2 - (3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-26+7i) \cdot (-4-3i)} {(-4)^2 - (3i)^2} = $ $ \dfrac{(-26+7i) \cdot (-4-3i)} {16 + 9} = $ $ \dfrac{(-26+7i) \cdot (-4-3i)} {25} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-26+7i}) \cdot ({-4-3i})} {25} = $ $ \dfrac{{-26} \cdot {(-4)} + {7} \cdot {(-4) i} + {-26} \cdot {-3 i} + {7} \cdot {-3 i^2}} {25} $ Evaluate each product of two numbers. $ \dfrac{104 - 28i + 78i - 21 i^2} {25} $ Finally, simplify the fraction. $ \dfrac{104 - 28i + 78i + 21} {25} = \dfrac{125 + 50i} {25} = 5+2i $